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a) Ta có: \(120^3-60\cdot120^2+1200\cdot120-7999\)
\(=120^3-3\cdot120^2\cdot20+3\cdot120\cdot20^2-20^3+1\)
\(=\left(120-20\right)^3+1\)
\(=100^3+1\)
\(=1000001\)
c) \(VT=\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3c\left(a+b\right)\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left[ab+c\left(a+b+c\right)\right]\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)
d) \(VT=a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\)
I don't now
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Ta có : \(\left(x-3\right)^3+3.\left(x+1\right)^2=\left(x^2-2x+4\right)\left(x+2\right)\)
\(\Leftrightarrow x^3-9x^2+27x-27+3.\left(x^2+2x+1\right)=x^3+8\)
\(\Leftrightarrow x^3-6x^2+33x-24=x^3+8\)
\(\Leftrightarrow-6x^2+33x-32=0\)
\(\Leftrightarrow6x^2-33x+32=0\)
\(\Leftrightarrow x=\frac{33\pm\sqrt{321}}{12}\)
Câu 9.
a) Ta có: \(\left(a-1\right)^2\ge0\)(điều hiển nhiên)
\(\Leftrightarrow a^2-2a+1\ge0\)
\(\Leftrightarrow a^2+2a+1\ge4a\)
\(\Leftrightarrow\left(a+1\right)^2\ge4a\left(đpcm\right)\)
b) Áp dụng BĐT Cauchy cho 2 số không âm:
\(a+1\ge2\sqrt{a}\)
\(b+1\ge2\sqrt{b}\)
\(c+1\ge2\sqrt{c}\)
\(\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\sqrt{abc}=8\)(Vì abc = 1)
Câu 10.
a) Ta có: \(-\left(a-b\right)^2\le0\)(điều hiển nhiên)
\(\Leftrightarrow-a^2+2ab-b^2\le0\)
\(\Leftrightarrow a^2+2ab+b^2\le2a^2+2b^2\)
\(\Leftrightarrow\left(a+b\right)^2\le2\left(a^2+b^2\right)\)
b) \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
Có: \(2ab\le a^2+b^2;2bc\le b^2+c^2;2ac\le a^2+c^2\)(BĐT Cauchy)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac\le3\left(a^2+b^2+c^2\right)\)
Vậy \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)
a) ( a - b - c )2 - ( a - b + c )2
= [ ( a - b - c ) - ( a - b + c ) ][ ( a - b - c ) + ( a - b + c ) ]
= ( a - b - c - a + b - c )( a - b - c + a - b + c )
= -2c( 2a - 2b )
= -2c.2( a - b )
= -4c( a - b )
b) ( a - x - y )3 - ( a + x - y )3
= [ ( a - x - y ) - ( a + x - y ) ][ ( a - x - y )2 + ( a - x - y )( a + x - y ) + ( a + x - y )2 ]
= ( a - x - y - a - x + y ){ [ ( a - x ) - y ]2 + [ ( a - y ) - x ][ ( a - y ) + x ] + [ ( a + x ) - y ] 2 }
= -2x{ [ ( a - x )2 - 2( a - x )y + y2 ] + [ ( a - y )2 - x2 ] + [ ( a + x )2 - 2( a + x )y + y2 ] }
= -2x{ [ a2 + x2 + y2 - 2ax - 2ay + 2xy ] + [ a2 - x2 + y2 - 2ay ] + [ a2 + x2 + y2 + 2ax - 2ay - 2xy ] }
= -2x{ a2 + x2 + y2 - 2ax - 2ay + 2xy + a2 - x2 + y2 - 2ay + a2 + x2 + y2 + 2ax - 2ay - 2xy }
= -2x{ 3a2 + x2 + 3y2 - 6ay } < trời ơi dài > ;-;